Question

[nguyen lan anh]

Cho a/b=c/d. Cmr

a, a²+b²/c²+ d² = ab/cd

b, (a+b)²/(c+d)² = ab/cd

Answer 1

[Nguyễn Huy Tú]

Giải:

Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)

a, Ta có: \(\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{b^2k^2+b^2}{d^2k^2+d^2}=\dfrac{b^2\left(k^2+1\right)}{d^2\left(k^2+1\right)}=\dfrac{b^2}{d^2}\) (1)

\(\dfrac{ab}{cd}=\dfrac{bkb}{dkd}=\dfrac{b^2}{d^2}\) (2)
Từ (1), (2) \(\Rightarrowđpcm\)

b, Ta có: \(\dfrac{\left(a+b\right)^2}{\left(c+d\right)^2}=\dfrac{\left(bk+b\right)^2}{\left(dk+d\right)^2}=\dfrac{\left[b\left(k+1\right)\right]^2}{\left[d\left(k+1\right)\right]^2}=\dfrac{b^2}{d^2}\) (1)

\(\dfrac{ab}{cd}=\dfrac{bkb}{dkd}=\dfrac{b^2}{d^2}\) (2)

Từ (1), (2) \(\Rightarrowđpcm\)

Answer 2

[Hoàng Thị Ngọc Anh]

Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)

\(\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\left(1\right)\)

a) Thay (1) vào đề:

\(VT=\dfrac{\left(bk\right)^2+b^2}{\left(dk\right)^2+d^2}=\dfrac{b^2k^2+b^2}{d^2k^2+d^2}=\dfrac{b^2\left(k^2+1\right)}{d^2\left(k^2+1\right)}=\dfrac{b^2}{d^2}\)

\(VP=\dfrac{bkb}{dkd}=\dfrac{b^2}{d^2}\)

\(\Rightarrow VT=VP\)

\(\Leftrightarrow\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{ab}{cd}\rightarrowđpcm.\)

b) Thay (1) vào đề bài:

\(\dfrac{\left(bk+b\right)^2}{\left(dk+d\right)^2}=\dfrac{\left[b\left(k+1\right)\right]^2}{\left[d\left(k+1\right)\right]^2}=\dfrac{b^2.\left(k+1\right)^2}{d^2.\left(k+1\right)^2}=\dfrac{b^2}{d^2}\)

Theo câu a) \(\dfrac{ab}{cd}=\dfrac{b^2}{d^2}\)

\(\Rightarrow\dfrac{\left(a+b\right)^2}{\left(c+d\right)^2}=\dfrac{ab}{cd}\rightarrowđpcm.\)